[.] is Greatest Integer Function
$$\lim_{x\to 0} \frac{-\tan( 10x^2) + \tan ^2 (10x^2)}{\sin^2 x}$$ $$=\lim_{x\to 0} \frac{(\tan 10x^2)(\tan 10 x^2 -1)}{\sin^2x}$$ $$=10 \lim_{x\to 0} \tan 10 x^2 -1$$ $$=-10$$
The right answer is $\tan 10 -10$
Can I please get my solution verified
Since $\lfloor -\pi^2\rfloor = -10$, the given function is $$\frac{\tan(-10x^2) - \tan^2(-10x^2)}{\sin^2 x} = \frac{-\tan 10x^2 - (-\tan 10x^2)^2}{\sin^2 x} = - \frac{(1 + \tan 10x^2)\tan 10x^2}{\sin^2 x}.$$ You did not correctly compute the second term in the numerator, forgetting that the negative is squared. However, this does not affect the value of the limit that you computed.
For your other question, note $$\frac{\tan(-10x^2) - \tan(-10) x^2}{\sin^2 x} = \frac{x^2}{\sin^2 x} \tan 10 - \frac{\sin(10x^2)}{\sin^2 x} \cdot \frac{1}{\cos(10x^2)}.$$ We know $(\sin x)/x \to 1$ as $x \to 0$. How would you evaluate $$\lim_{x \to 0} \frac{\sin (10x^2)}{\sin^2 x}?$$ There are multiple ways to do this. I leave this to you as an exercise.